m
v
p
Honors
MOL
Chemistry
Unit 6:
The Mathematics of
Chemical Formulas
Cu(OH)2
# of H2O
molecules
# of H atoms
# of O atoms
1
2
1
2
4
2
3
6
3
100
200
100
6.02 x 1023
2 (6.02 x 1023)
6.02 x 1023
18.0 g
2.0 g
16.0 g
molar mass: the mass of one mole of a substance
PbO2
HNO3
Pb:
1 (207.2 g) = 207.2 g
O:
2 (16.0 g) = 32.0 g
H:
1 (1.0 g)
N:
1 (14.0 g) = 14.0 g
O:
3 (16.0 g) = 48.0 g
ammonium phosphate
(NH4)3PO4 N:
239.2 g
= 1.0 g
63.0 g
NH4+ PO43–
3 (14.0 g) = 42.0 g
H: 12 (1.0 g)
= 12.0 g
P:
1 (31.0 g) = 31.0 g
O:
4 (16.0 g) = 64.0 g
149.0 g
percentage composition:
the mass % of each element in a compound
g element
x 100
% of element =
molar mass of compound
Find % composition
PbO2
239.2 g
(NH4)3PO4
149.0 g
(see calculation above)
207.2 g Pb : 239.2 g = 86.6% Pb
32.0 g O : 239.2 g = 13.4% O
42.0 g N
12.0 g H
31.2 g P
64.0 g O
: 149.0 g
: 149.0 g
: 149.0 g
: 149.0 g
=
=
=
=
28.2% N
8.1% H
20.8% P
43.0% O
zinc acetate
CH3COO–
Zn2+
Zn(CH3COO)2
Zn: 1 (65.4 g) = 65.4 g
= 35.7% Zn
C: 4 (12.0 g) = 48.0 g
= 26.2% C
H: 6 (1.0 g) =
6.0 g
O: 4 (16.0 g) = 64.0 g
183.4 g
: 183.4 g
= 3.3% H
= 34.9% O
Mole Calculations
Must Use Periodic Table!
Mass
(g)
Volume
(L or dm3)
1 mol = molar mass (in g)
1 mol = 22.4 L
1 mol = 22.4 dm3
Particle
(at. or
m’c)
MOLE
(mol)
1 mol = 6.02 x 1023 particles
1 mol
Remember…we use the conversions
to set up ratios and cancel units
6.02 x 1023 OR
particles
6.02 x 1023 particles
1 mol
1 mol = molar mass (in g)
Mass
(g)
Volume
(L or dm3)
New Points about
Island Diagram:
1 mol = 22.4 L
MOLE
(mol)
1 mol = 22.4 dm3
Particle
(at. or
m’c)
1 mol = 6.02 x 1023
particles
a. Diagram now has four islands
b. “Mass Island” now for elements or compounds
c. “Particle Island” now for atoms or molecules
d. “Volume Island”: for gases only
1 mol @ STP = 22.4 L = 22.4 dm3
1. What mass is 1.29 mol iron (II) nitrate ?
Fe2+ NO3–
Fe(NO3)2
1.29 mol 179.8 g
1 mol
= 232 g
2. How many molecules is 415 L sulfur dioxide at STP?
SO2
415 L
1 mol
22.4 L
6.02 x 1023 m’c
1 mol
= 1.12 x 1025 m’c
3.What mass is 6.29 x 1024 m’cules aluminum sulfate ?
Al3+
SO42–
Al2(SO4)3
6.29 x 1024 m’c
1 mol
6.02 x 1023 m’c
342.3 g
1 mol
= 3580 g
4. At STP, how many g is 87.3 dm3 of nitrogen gas?
N2
87.3 L
1 mol
28.0 g
22.4 L
1 mol
= 109 g
5. How many m’cules is 315 g of iron (III) hydroxide?
Fe3+
OH–
Fe(OH)3
315 g
1 mol
106.8 g
6.02 x 1023 m’c
1 mol
= 1.78 x 1024 m’c
6. How many atoms are in 145 L of CH3CH2OH at STP?
1 mol 6.02 x 1023 m’c
1 mol
22.4 L
= 3.90 x 1024 m’c
But there are 9 atoms per molecule, so…
145 L
9 (3.90 x 1024) = 3.51 x 1025 atoms
Finding an Empirical Formula from Experimental Data
a. Find # of g of each element.
“What’s your flavor
of ice cream?”
b. Convert each g to mol.
c. Divide each “# of mol” by the smallest “# of mol.”
d. Use ratio to find formula.
1. A compound is 45.5% yttrium and 54.5% chlorine.
Find its empirical formula.
 1 mol Y 
  0.512 mol Y  0.512  1
45.5 g Y 
 88.9 g Y 
 1 mol Cl 
  1.535 mol Cl  0.512  3
54.5 g Cl 
 35.5 g Cl 
YCl3
2. A ruthenium/sulfur
compound is 67.7% Ru.
Find its empirical formula.
 1 mol Ru 
  0.670 mol Ru  0.670  1
67.7 g Ru 
 101.1 g Ru 
 1 mol S 
  1.006 mol S  0.670  1.5
32.3 g S 
 32.1 g S 
RuS1.5
Ru2S3
3. A 17.40 g sample of a
technetium/oxygen compound
contains 11.07 g of Tc. Find
the empirical formula.
 1 mol Tc 
  0.113 mol Tc  0.113  1
11.07 g Tc 
 98 g Tc 
 1 mol O 
  0.396 mol O  0.113  3.5
6.33 g O 
 16.0 g O 
TcO3.5
Tc2O7
4. A compound contains 4.63 g lead, 1.25 g nitrogen,
and 2.87 g oxygen. Name the compound.
 1 mol Pb 
  0.0223 mol Pb  0.0223  1
4.63 g Pb 
 207.2 g Pb 
 1 mol N 
  0.0893 mol N  0.0223  4
1.25 g N 
 14.0 g N 
 1 mol O 
  0.1794 mol O  0.0223  8
2.87 g O 
 16.0 g O 
?
PbN4O8
?
Pb(NO2)4
Pb? 4 NO2–
lead (IV) nitrite
(plumbic nitrite)
To find molecular formula…
a. Find empirical formula.
b. Find molar mass of
empirical formula.
c. Find n = mm molecular
mm empirical
d. Multiply all parts of
empirical formula by n.
(“What’s your
flavor?”)
(“How many scoops?”)
(How many empiricals “fit into” the molecular?)
1. A carbon/hydrogen compound is 7.7% H and has a
molar mass of 78 g. Find its molecular formula.
 1 mol H 
  7.7 mol H  7.69  1
7.7 g H 
 1.0 g H 
 1 mol C 
  7.69 mol C  7.69  1
92.3 g C 
 12.0 g C 
emp. form.  CH
mmemp = 13 g
78 g
=6
13 g
C6H6
2. A compound has 26.33 g nitrogen, 60.20 g oxygen,
and molar mass 92 g. Find molecular formula.
 1 mol N 

26.33 g N 
 14.0 g N 
 1.881mol N  1.881  1
 1 mol O 
  3.763 mol O  1.881  2
60.20 g O 
 16.0 g O 
NO2
mmemp = 46 g
92 g
=2
46 g
N2O4
Hydrates and Anhydrous Salts
anhydrous salt: an ionic compound (i.e., a salt) that
attracts water molecules and forms
loose chemical bonds with them;
symbolized by MN
“anhydrous” = “without water”
Uses: “desiccants” in leather
MN = metal +
goods, electronics, vitamins
nonmetal
hydrate: an anhydrous salt with the water attached
symbolized by MN . ? H2O
Examples: CuSO4 . 5 H2O
Na2CO3 . 10 H2O
BaCl2 . 2 H2O
FeCl3 . 6 H2O
H 2O H 2O
H2O MN H2O
H 2O H 2O
hydrate
HEAT
MN
H 2O
H 2O
HO
H 2O H O 2
H 2O
2
+
anhydrous salt
ENERGY
ENERGY
water
+
+
Finding the Formula
of a Hydrate
1. Find the # of g of MN and # of g of H2O
2. Convert g to mol
3. Divide each “# of mol” by the smallest “# of mol”
4. Use the ratio to find the hydrate’s formula
Find formula of hydrate for each problem
H2O
sample’s mass before
heating = 4.38 g
MN . ? H2O
(hydrate)
sample’s mass after
MN
heating = 1.93 g (anhydrous salt)
molar mass of anhydrous salt = 85 g
 1 mol MN 
  0.0227 mol MN  0.0227  1
1.93 g MN 
 85 g MN 
 1 mol H2O 
  0.1361 mol H2O  0.0227  6
2.45 g H2O
 18 g H2O 
MN . 6 H2O
A. beaker = 46.82 g
MN
H2O
B. beaker + sample before
heating = 54.35 g
C. beaker + sample after
heating = 50.39 g
beaker +
salt + water
beaker + salt
molar mass of anhydrous salt = 129.9 g
 1 mol MN 
  0.0275 mol MN  0.0275  1
3.57 g MN 
 129.9 g MN 
 1 mol H2O 
  0.22 mol H2O  0.0275  8
3.96 g H2O 
 18 g H2O 
MN . 8 H2O
A. beaker = 47.28 g
MN
H2O
B. beaker + sample before
heating = 53.84 g
C. beaker + sample after
heating = 51.48 g
beaker +
salt + water
beaker + salt
molar mass of anhydrous salt = 128 g
 1 mol MN 
  0.0328 mol MN  0.0328  1
4.20 g MN 
 128 g MN 
 1 mol H2O 

2.36 g H2O 
 18 g H2O 
 0.0328  4
MN . 4 H2O
For previous problem, find % water and
% anhydrous salt (by mass).
g H2O in formula
% H2O 
molar mass of hydrate
4 (18 g H2O)
 36.0% H2O
% H2O 
4 (18 g )  128 g
64.0% MN
or…
2.36 g H2O
% H2O 
 36.0% H2O
2.36 g  4.20 g
64.0% MN
Review Problems
1. Find % comp. of iron (III) chloride.
Fe3+
Cl–
FeCl3
Fe: 1 (55.8 g) = 55.8 g
Cl: 3 (35.5 g) = 106.5 g
162.3 g
: 162.3 g
 34.4% Fe
 65.6% Cl
2. A compound contains 70.35 g C and 14.65 g H.
Its molar mass is 58 g. Find its molecular formula.
 1 mol H 
  14.65 mol H  5.86  2.5
14.65 g H 
 1.0 g H 
 1 mol C 
  5.86 mol C  5.86  1
70.35 g C 
 12.0 g C 
emp. form.  C2H5
mmemp = 29 g
58 g
=2
29 g
C4H10
3. At STP, how many g is
548 L of chlorine gas?
Cl2
548 L
(
1 mol
22.4 L
)(
71.0 g
1 mol
)=
1740 g
4.Strontium chloride is an anhydrous salt on which the
following data were collected. Find formula of hydrate.
A. beaker = 65.2 g
B. beaker + sample before
heating = 187.9 g
C. beaker + sample after
heating = 138.2 g
Sr2+
Cl1–
SrCl2
beaker +
salt + water
beaker + salt
 1 mol MN 
  0.46 mol MN  0.46  1
73.0 g MN 
 158.6 g MN 
 1 mol H2O 
  2.76 mol H2O  0.46  6
49.7 g H2O 
 18 g H2O 
SrCl2 . 6 H2O